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Find all positive integer solutions of $a^3 + 2b^3 = 4c^3$.

Proof: There don't exist any integer solutions for the give equation. Proof by the Well Ordering Principle.

Let $d$ be the set of all such possible combinations in form $(a, b, c)$.

By the Well Ordering Principle, we that every nonempty set of non-negative integers always has a smallest element. So, there must be a tuple $(a, b, c)$ such that $\max(a, b, c)$ is the least possible value in $d$.

We also know that $a$, $b$ and $c$ must be even. Since $a$ must be even if $4c^3$ has to be even but we know that $4c^3$ is even, so $a$ is the same. Similarly it can be shown that all $a$, $b$ and $c$ are even.

Now, let $a = 2x$, $b = 2y$ and $c = 2z$.

But then, $$a^3 + 2b^3 = 4c^3$$ $$\implies (2x)^3 + 2*(2y)^3 = 4*(2z)^3$$ $$\implies 8x^3 + 16y^3 = 32z^3$$ $$\implies x^3 + 2y^3 = 4z^3$$

So, $(x, y, z) \in d$. But, $\max(x, y, z) < \max(a, b, c)$. But $(a, b, c)$ was supposed to be the smallest such member. We have a contradiction here. There is no smallest $(a, b, c)$ which satisfy this criterion and hence no $(a, b, c)$ at all and $d$ is empty. Hence, there exist no integers $a$, $b$ and $c$ which satisfy $a^3 + 2b^3 = 4c^3$

I wrote this proof. Is it enough to prove the above question? Am I right? Are there any mistakes in the above proof? If there are any, please help me.

Do you have any other ways of proving it?

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  • $\begingroup$ I can't see any mistakes; looks like a superb proof. It took me a while to understand your reasoning why $b$ and $c$ must be even, but I mean it is true (because $2^3 > \max\{4,2\}$, in essence!). Just to let you know, there is some tag like "proof-verification" that you might want to add. But yeah, +1 for an excellent question, showing you've really thought about it. $\endgroup$
    – Sam OT
    Feb 20 '16 at 10:16
  • $\begingroup$ Related : math.stackexchange.com/questions/422440/… $\endgroup$ Feb 20 '16 at 13:54
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There is another proof using the well ordering principle.

We use the following fact:

Third powers are can only be congruent to $0 $, $1 $ or $8 \mod 9 $.

Now, we see that the left hand side is congruent to $0$, $1$, $2$, $3$, $6$, $7$ or $8 \mod 9$, while the right hand side is congruent to $0, 4$ or $5 \mod 9$. So the only possibility is that $4c^3 \equiv 0 \mod 9$ and $a^3+2b^3 \equiv 0 \mod 9$. The first modular congruence gives that $3 \mid c$. The second can only hold true if $3 \mid a$ and $3 \mid b$.

If one of those isn't true, then we get either a combination of $0 \mod 9$ and $2$ or $7 \mod 9$, a combination of $0 \mod 9$ and $1$ or $8 \mod 9$, a combination of $1$ or $8 \mod 9$ and $2$ or $7 \mod 9$, none of which add to $0 \mod 9$.

Now suppose that $(a,b,c)$ is the solution with the smallest $a$.

We know that $3 \mid c$, $3 \mid a$ and $3 \mid b$.

Hence write $a=3x, b=3y, c=3z$ and then is $(x,y,z)$ also a solution with smaller $a$.

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  • $\begingroup$ Why did you claim that $a$, $b$, $c$ are congruent to 0 mod 7? $\endgroup$ Feb 20 '16 at 10:26
  • $\begingroup$ @DhruvSomani It was a typo, it was meant to be 0 mod 9. I will rewrite the answer, because now there are more mistakes. $\endgroup$
    – wythagoras
    Feb 20 '16 at 11:08

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